How do you solve $(8sqrt( 2x-1)) / 3 = x$ ?

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Answer 1 · 2016-02-16T02:44:40

First, send the 3 to the other side in multiplication.

$8(sqrt(2x - 1)) = 3x$

$sqrt(2x - 1) = 3/8x$

Get rid of the square root by squaring both sides of the equation.

$(sqrt(2x - 1))^2 = (3/8x)^2$

$2x - 1 = 9/64x^2$

$64(2x - 1) = 9x^2$

$128x - 64 = 9x^2$

$-64 = 9(x^2 + 128/9 + m - m)$

$m = (b/2)^2$

$m = ((128/9)/2)^2$

$m = 16384/324$

$-64 = 9(x^2 + 128/9 + 16384/324 - 16384/324)$

$-64 = 9(x^2 + 128/9 + 16384/324) - 147456/324$

$(-64 + 147456/324)/9 = (x + 128/18)^2$

$126720/2916 = (x + 128/18)^2$

$+-sqrt(126720/2915) - 128/18 = x$

$-0.52 ~= x and -13.70 ~= x$

When you plug both these answers into the original equation, they both make the square root negative inside. This means there is no solution

Hopefully this helps!

How do you solve (8sqrt( 2x-1)) / 3 = x(8sqrt( 2x-1)) / 3 = x?